Prove by mathematical induction that $31\mid5^{n+1}+6^{2n-1},\;$ $\forall n \in \mathbb{N}$

Question

The proof of the statement $\mathcal{S}(n)=31\mid5^{n+1}+6^{2n-1}$ is going to be conducted by mathematical induction in terms of $n\in\mathbb{N}.$

1. Base clause

Show that $\mathcal{S}(1)$ holds.

$$\begin{align}\mathcal{S}(1)&=31\mid5^{1+1}+6^{2\cdot1-1}\\&=31\mid5^2+6\\&=31\mid25+6\\&=31\mid31\;\checkmark\end{align}$$

2. Induction hypothesis

Assume that $\mathcal{S}(n)$ holds for any fixed but arbitrary $n\in\mathbb{N}$.

3. Induction step

Show that $\mathcal{S}(n+1)$ holds. $$ \begin{align} \mathcal{S}(n+1) & = 31\mid5^{(n+1)+1}+6^{2(n+1)-1} \\ & = 31\mid5^{n+2}+6^{2n+1} \\ & = 31\mid5^{n+1}\cdot5+6^{2n-1}\cdot6^2 \end{align} $$

At this point I don't know how to proceed. I really do have the feeling of missing something terribly obvious.

I tried to recreate the terms of the sum in the original statement $\mathcal{S}(n)$ with the goal of figuring out how to apply the induction hypothesis.

At this point I am also aware that $$\mathcal{S}(n)\Leftrightarrow \exists\varphi\in\mathbb{N}:31\varphi=5^{n+1}+6^{2n-1}$$ and therefore $$\mathcal{S}(n+1)\Leftrightarrow \exists\varphi\in\mathbb{N}:31\varphi=5^{n+1}\cdot5+6^{2n-1}\cdot6^2$$

Answer 1

HINT: $$5^{n+2}+6^{2n+1}=5\cdot5^{n+1}+\underbrace{36}_{=5+31}\cdot6^{2n-1}=5\cdot(5^{n+1}+6^{2n-1})+31\cdot 6^{2n-1}$$

Answer 2

From the induction hypothesis, $$31\mid (5^{n+1}+6^{2n+1})(5+6^2),$$ or $$31\mid 5^{(n+1)+1}+6^{2(n+1)+1}+5\cdot 6^{2n+1}+6^2\cdot 5^{n+1}.$$ So you are left to show $$31\mid 5\cdot 6^{2n+1}+6^2\cdot 5^{n+1}.$$ Applying the induction hypothesis again, you have $$31\mid 6^2\cdot 6^{2n+1}+6^2\cdot 5^{n+1}.$$ So you are left with $$31\mid (6^2-5)6^{2n+1},$$ which is true.

Answer 3

First, show that this is true for $n=1$:

$6^{2-1}+5^{1+1}=31$

Second, assume that this is true for $n$:

$6^{2n-1}+5^{n+1}=31k$

Third, prove that this is true for $n+1$:

$6^{2n+1}+5^{n+2}=$

$36\cdot6^{2n-1}+5\cdot5^{n+1}=$

$36\cdot6^{2n-1}+(36-31)\cdot5^{n+1}=$

$36\cdot6^{2n-1}+36\cdot5^{n+1}-31\cdot5^{n+1}=$

$36\cdot(\color\red{6^{2n-1}+5^{n+1}})-31\cdot5^{n+1}=$

$36\cdot\color\red{31k}-31\cdot5^{n+1}=$

$31\cdot(36k-5^{n+1})$

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Please note that the assumption is used only in the part marked red.

Answer 4

Hint $\ $ The inductive step follows very simply by using $\,\rm{\color{#C00}{CPR}} = $ Congruence Product Rule to multiply the first two congruences below

$$\begin{align} {\rm mod}\,\ 31\!:\qquad\ 5\,\ &\equiv\,\ 6^{\large 2}\\[0.3em] 5^{\large N+1}&\equiv -6^{\large 2N-1}\ \ \ {\rm i.e.}\ \ P(N)\\ \overset{\rm{\color{#C00}{CPR}}}\Longrightarrow\ \ \ 5^{\large N+2}&\equiv -6^{\large 2N+1}\ \ \ {\rm i.e.}\ \ P(N\!+\!1) \end{align}$$

Remark $\ $ The common inductive proofs using divisibility in other answers essentially do the same thing, i.e. they repeat the proof of the Congruence Product Rule in this special case, but expressed in divisibility vs. congruence language (e.g. see here). But unfortunately the product rule is much less arithmetically intuitive when expressed as unstructured divisibilities, which greatly complicates the discovery of the inductive step. I explain this at length in other answers, e.g. see here.